Let $\pi: V\to S$ be a standard conic bundle of a threefold $V$ to a surface $S$, i.e., $\pi$ is relative minimal. Assume that everything is nonsingular and is over $\mathbb{C}$. We may assume that $V$ is embedded in a $\mathbb{P}^2$-bundle $\mathbb{P}(\mathcal{E})$ over $S$, where $\mathcal{E}$ is a rank $3$ vector bundle on $S$, and $V$ is defined as a zero of a section $\sigma\in H^0(\mathbb{P}(E),\mathcal{O}_{\mathbb{P}(E)}(2)\otimes\tau^*\mathcal{L})$, where $\mathcal{L}$ is an invertible sheaf on $S$ and $\tau$ is the standard projection of $\mathbb{P}(\mathcal{E})$ to $S$.

Now suppose that $C$ is a curve on $V$ which is isomorphic to its image $\pi(C)$, such that $\pi(C)\cap\Delta=\varnothing$, where $\Delta$ is the degenerate divisor on $S$ for the conic bundle $(V, S,\pi)$, that is the locus of points whose fiber is a degenerate conic, disjoint of two lines or a double line. Then we can apply an elementary transformation to $V$, as described in this paper On conic bundle structures--V. G. Sarkisov, which first blows up $C$ then blows down the strict transform of $B=\pi^{-1}(\pi(C))$. The resulting variety is denoted by $V'$, and $C'$ is the strict transform of $C$.

Question: Is this elementary transformation a flop or a flip in the sense of the MMP?

I try to answer this question positively, in other words, does the intersection numbers hold $K_V.C<0$ and $K_{V'}.C'>0$? But I meet some trouble when I verify this inequality, by 1.5 of On conic bundle structures--V. G. Sarkisov, we know that $\mathcal{L}$ is isomorphic to $\omega_S\otimes\mathrm{det}(\mathcal{E})$, so $\omega_{\mathbb{P}(\mathcal{E})}\otimes\mathcal{O}_{\mathbb{P}(\mathcal{E})}(V)\cong\mathcal{O}_{\mathbb{P}(\mathcal{E})}(-1)$. Therefore the intersection number of $K_V.C$ should be equal to $(K_{\mathbb{P}(\mathcal{E})}+V).C=\mathrm{deg}_C\mathcal{O}_{\mathbb{P}(\mathcal{E})}(-1)$, as $C$ contained in $V$. We only konw that the divisor associate to $\mathcal{O}_{\mathbb{P}(\mathcal{E})}(-1)$ is relatively ample, but here the curve $C$ is not in a fiber, we can not conclude that $\mathrm{deg}_C\mathcal{O}_{\mathbb{P}(\mathcal{E})}(-1)<0$ directly.

On the other hand, it seems that the situation for $(V',S,\pi')$ is the same as $(V,S,\pi)$. Since $(V',S,\pi')$ is also standard. So we have $K_{V'}.C'=\mathrm{deg}_{C'}\mathcal{O}_{\mathbb{P}(\mathcal{E})}(-1)$. Note that $B$ (resp. $B'=\pi'^{-1}(\pi'(C'))$) is a ruled surface over $\pi(C)$ (resp. $\pi'(C)$), if my question has a positive anwser, it seems that $C$ is the negative section of $B\to\pi(C)$ and $C'$ is a positive section of $B'\to \pi'(C')$, i.e., $C^2<0$ and $C'^2>0$. Let $W\to V$ be the blowing up along $C$, and let $\tilde{B}$ and $\tilde{B'}$ be the strict transforms of $B$ and $B'$ respectively. It seems that $\tilde{B}$ and $\tilde{B'}$ are guling along a positive section of $\tilde{B}$ with a negative section of $\tilde{B'}$ in $W$. But I don't know how to prove this. However, it may be false.

  • 2
    $\begingroup$ It is not a flip or flop (since there is no extremal contraction of the curve $C\subset V$ in this setting). Instead it is an example of a Sarkisov link (see, eg., Hacon & McKernan, `The Sakisov Program'). $\endgroup$
    – Tom Ducat
    Apr 12, 2021 at 10:57

1 Answer 1


It is neither flip nor flop, because the exceptional locus both on $V$ and on $V'$ is a divisor.


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